In this scenario, for each node that is a momentum eigenstate, it imparts a random displacement – sampled from the uniform distribution of its \(q\) -quadrature since we assumed \(\delta \rightarrow 0\)  – onto the \(p\) -quadrature of its neighbors via the action of the CV \(CZ\)  gates, while each node that is a perfect GKP state does not impart any random displacements onto its neighbors. Let the displacements from \(e_1,e_2,e_3,e_4\)  be \(d_1,d_2,d_3,d_4\) , where we specifically mean the excess displacement beyond \(n\sqrt{\pi }\)  [1]}. Thus, the displacement on \(e_0\)  is given by \(d_1+d_2+d_3+d_4\) , assuming the \(CZ\)  gates are all +1; changing the sign of the \(CZ\)  gates does not change the argument since \(d_1,d_2,d_3,d_4\)  are sampled from symmetric distributions. Moreover, let \(b_i,\,i=1,2,3,4\)  be the binary value returned by performing standard GKP binning on the homodyne output of the neighbors of node \(e_i\)  other than \(e_0\) ; note that the only shared neighbor of \(e_1,e_2,e_3,e_4\)  is \(e_0\) , and since we assumed the nodes beyond \(e_1,e_2,e_3,e_4\)  were GKP, then we know we will get the same singular outcome \(b_i\)  on all neighbors of \(e_i\)  other than \(e_0\) . Let \(b_0\)  be the binary value returned by performing standard GKP binning on the homodyne output of \(e_0\) .
